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Year 10 Interactive Maths - Second Edition


Elimination Method

To solve the simultaneous equations, make the coefficients of one of the variables the same value in both equations. Then either add the equations or subtract one equation from the other (whichever is appropriate) to form a new equation that only contains one variable. This is referred to as eliminating the variable.

Solve the equation thus obtained. Then substitute the value found for the variable in one of the given equations and solve it for the other variable. Write the solution as an ordered pair.


Example 4

Solve the following simultaneous equations by using the elimination method:

2x + 3y = 15 and 4x - 3y = 3

Solution:

Label the equations as follows:

2x + 3y = 15   ...(1)     and     4x - 3y = 3   (2)

Notice that 3y appears on the left-hand side of both equations.  Adding the left-hand side of (1) and (2), and then the right-hand sides, gives:

2x + 3y + 4x - 3y = 15 + 3, so we find x = 3   

Note:

We have added equals to equals, and addition eliminates y.


Substituting x = 3 in (1) gives:

2(3) + 3y = 15, so we find y = 3

So, the solution is (3, 3).


Example 5

Solve the following simultaneous equations by using the elimination method:

11x + 6y = 79 and 11x + 3y = 67

Solution:

Label the equations as follows:

11x + 6y = 79   ...(1)     and     11x + 3y = 67   ...(2)

Subtracting (2) from (1) gives:

11x + 6y - 11x - 3y = 79 - 67, so we find y = 4   

Substituting y = 4 in (1) gives:

11x + 6(4) = 79, so we find x = 5

So, the solution is (5, 4)


Example 6

Solve the following simultaneous equations by using the elimination method:

2x + 3y = 13     and     3x + 2y = 12

Solution:

Label the equations as follows:

2x + 3y = 13   ...(1)     and     3x + 2y = 12   ...(2)

Multiplying (1) by 2 and (2) by 3 gives:

4x + 6y = 26   ...(3)     and     9x + 6y = 36   ...(4)   

Subtracting (3) from (4) gives:

9x + 6y - 4x - 6y = 36 - 26, so x = 2

Substituting x = 2 in (1) gives 2(2) + 3y = 13, so we find y = 3

So, the solution is (2, 3).


Equations Containing Fractions

If either or both equations contain fractions, then it is desirable to remove the denominators before solving them simultaneously.

Example 7

Solve the following simultaneous equations by using the elimination method:

x / 2 + y / 3 = 6   and   2x - y = 3

Solution:

Label the equations as follows:

x / 2 + y / 3 = 6   ...(1)     and     2x - y = 3   ...(2)

To simplify equation (1), multiply both sides by 6, the lowest common denominator of 2 and 3.

3x + 2y = 36   ...(3)        {Equation 3 replaces equation 1}

The equations are now:

3x + 2y = 36   ...(3)         2x - y = 3   ...(2)

Multiplying (2) by 2 gives:

4x - 2y = 6   ...(4)     {Equation (4) replaces (2)}

Adding (3) and (4) gives:

7x = 42, so x = 6

Substituting x = 6 in (2) gives:

2(6) - y = 3 so y = 9

So, the solution is (6, 9).


Key Terms

elimination method, ordered pair


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